It is evident that if we assume that \(\Phi \left( n\right)\ge cn,\)  where \(c\)  is some positive constant then
\(\frac{\log ^p n\Phi \left(n\right)}{n^{2-2p}}\ge n^{1-2p}\log ^p n \rightarrow \infty , \ \text{as} \ n\rightarrow \infty ,\) 
and also (REF ) holds. So, without lost the generality we may assume that there exists an increasing sequence of positive integers
\(\left\lbrace  \alpha ^{\prime } _{k}:k\in \mathbb {N} \right\rbrace  \)  such that
\( \Phi \left( \alpha ^{\prime } _{k}\right)=o(\alpha ^{\prime } _{k}), \ \text{as} \ k\rightarrow \infty .\) 
Let \(\left\lbrace  \alpha _{k}:k\in \mathbb {N} \right\rbrace \subseteq \left\lbrace  \alpha ^{\prime } _{k}:k\in \mathbb {N} \right\rbrace \)  be an increasing sequence of positive integers such that \(\alpha _{0}\ge 2\)  and
\(\sum _{k=0}^{\infty }\frac{1}{\Phi ^{1/2}(2^{2\alpha _k})}<\infty , \) 
\( \sum _{\eta =0}^{k-1}\frac{2^{2\alpha _{\eta }/p}}{\Phi ^{1/2p}(2^{2\alpha _\eta })}\le \frac{2^{2\alpha _{k-1}/p+1}}{\Phi ^{1/2p}(2^{2\alpha _{k-1}})},\) 
\( \frac{2^{2\alpha _{k-1}/p+1}}{\Phi ^{1/2p}(2^{2\alpha _{k-1}})}\le \frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}.\) 
We note that under condition (REF ) we can conclude that
\(\frac{2^{2\alpha _{\eta }/p}}{\Phi ^{1/2p}(2^{2\alpha _\eta })}\ge {\left(\frac{2^{2\alpha _{\eta }}}{\Phi (2^{2\alpha _\eta })}\right)}^{1/2p} \rightarrow \infty , \ \text{as} \ \eta \rightarrow \infty \) 
and it immediately follows that such an increasing sequence \(\left\lbrace  \alpha _k:k\in \mathbb {N}\right\rbrace  ,\)  which satisfies conditions (REF )-(REF ), can be constructed.
Let
\(f^{\left( A\right) }\left( x\right) :=\sum _{\left\lbrace  k;\text{ }2\alpha _{k}<A\right\rbrace  }\lambda _{k}a_{k},\) 
where
\(\lambda _k=\frac{1}{\Phi ^{1/2p}(2^{2\alpha _k})}\) 
and
\(a_{k}={2^{2\alpha _{k}(1/p-1)}}\left( D_{2^{2\alpha _{k}+1}}-D_{2^{2\alpha _{k}}}\right) .\) 
From (REF ) and Lemma REF  we can conclude that \(f=\left( f^{\left(n\right) },n\in \mathbb {N} \right) \in H_p(G_2).\)
It is easy to show that
\( \widehat{f}^w(j)=\left\lbrace \begin{array}{l}\frac{2^{2\alpha _{k}(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})},\,\,\text{ if \,\,}j\in \left\lbrace  2^{2\alpha _{k}},...,\text{ ~}2^{2\alpha _{k}+1}-1\right\rbrace  ,\text{ }k\in \mathbb {N}, \\0,\text{ \,\,\,if \,\,\,}j\notin \bigcup \limits _{k=1}^{\infty }\left\lbrace  2^{2\alpha _k},...,\text{ ~}2^{2\alpha _{k}+1}-1\right\rbrace .\end{array}\right.\) 
For \(n=\sum _{i=1}^s 2^{n_i},\)  \(n_1<n_2<...<n_s\)  we denote
\(\mathbb {A}_{0,2}:=\left\lbrace  n\in \mathbb {N}:\text{ }n=2^0+2^2+\sum _{i=3}^{s_n}2^{n_i}\right\rbrace .\) 
Let \(2^{2\alpha _{k}}\le j\le 2^{2\alpha _{k}+1}-1\)  and \(j\in \mathbb {A}_{0,2}.\)  Then
\( R^w_jf &=&\frac{1}{l_j}\sum _{n=1}^{2^{2\alpha _k}-1}\frac{S_nf}{n} +\frac{1}{l_j}\sum _{n=2^{2\alpha _k}}^{j}\frac{S_{n}f}{n}:=I+II.\) 
Let \(n<2^{2\alpha _{k}}.\)  Then from (REF ), (REF ) and (REF ) we have that
\(\left|S^w_{n}f\left( x\right) \right|&\le &\sum _{\eta =0}^{k-1}\sum _{v=2^{2\alpha _{\eta }}}^{2^{2\alpha _{\eta }+1}-1}\left|\widehat{f}^w(v)\right|\le \sum _{\eta =0}^{k-1}\sum _{v=2^{2\alpha _{\eta }}}^{2^{2\alpha _{\eta }+1}-1}\frac{2^{2\alpha _{\eta }(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _\eta })} \\&\le &\sum _{\eta =0}^{k-1}\frac{2^{2\alpha _{\eta }/p}}{\Phi ^{1/2p}(2^{2\alpha _\eta })}\le \frac{2^{2\alpha _{k-1}/p+1}}{\Phi ^{1/2p}(2^{2\alpha _{k-1}})}\le \frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}.\) 
Consequently,
\( \left|I\right|&\le &\frac{1}{l_j}\underset{n=1}{\overset{2^{2\alpha _k}-1}{\sum }}\frac{\left|S^w_n f\left( x\right)\right|}{n} \\&\le &\frac{1}{l_{2^{2\alpha _k}}}\frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\sum _{n=1}^{2^{2\alpha _k}-1} \frac{1}{n}\le \frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}. \) 
Let \(2^{2\alpha _{k}}\le n\le 2^{2\alpha _{k}+1}-1.\)  Then we have the
following
\(S^w_{n}f &=&\sum _{\eta =0}^{k-1}\sum _{v=2^{2\alpha _{\eta }}}^{2^{2\alpha _{\eta }+1}-1}\widehat{f}^w(v)w_{v}+\sum _{v=2^{2\alpha _{k}}}^{n-1}\widehat{f}^w(v)w_{v} \\&=&\sum _{\eta =0}^{k-1}\frac{2^{{2\alpha _{\eta }}\left(1/p-1\right)}}{\Phi ^{1/2p}(2^{2\alpha _\eta })}\left( D^w_{2^{2\alpha _{\eta }+1}} -D^w_{2^{2\alpha _{\eta }}}\right)+\frac{2^{{2\alpha _k}\left( 1/p-1\right)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left(D^w_{n}-D^w_{2^{2\alpha _k}}\right) .\) 
This gives that
\(II=\frac{1}{l_j}\underset{n=2^{2\alpha _k}}{\overset{2^{2\alpha _{k}+1}}{\sum }}\ \frac{1}{n}\left(\sum _{\eta =0}^{k-1}\frac{2^{2\alpha _{\eta }\left(1/p-1\right)}}{\Phi ^{1/2p}(2^{2\alpha _\eta })}\left( D^w_{2^{2\alpha _{\eta }+1}}-D^w_{2^{2\alpha _{\eta }}}\right)\right)\) 
\(+\frac{1}{l_{j}}\frac{2^{2\alpha _k\left(1/p-1\right)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\sum _{n=2^{2\alpha _k}}^{j}\frac{\left( D^w_n-D^w_{2^{2\alpha _k}}\right) }{n}:=II_{1}+II_{2}.\) 
Let \(x\in I_{2}(e_{0}+e_{1})\in I_{0}\backslash I_{1}.\)  According to well-known equalities for Dirichlet kernels (for details see e.g. [1]} and [2]}):
recall that
\( D^w_{2^{n}}\left( x\right)=\left\lbrace \begin{array}{ll}2^{n}, & \,\text{if\,\,\,}x\in I_{n} \\0, & \text{if}\ \ \ x\notin I_{n}\end{array}\right.\) 
and
\(D^w_n=w_n\overset{\infty }{\underset{k=0}{\sum }}n_kr_kD^w_{2^k}=w_n\overset{\infty }{\underset{k=0}{\sum }}n_{k}\left(D^w_{2^{k+1}}-D^w_{2^{k}}\right),\text{ for \ }n=\overset{\infty }{\underset{i=0}{\sum }}n_{i}2^{i},\) 
so we can conclude that
\(D^w_n\left( x\right) =\left\lbrace \begin{array}{ll}w_{n}, & \,\text{if\,\,\,}n\ \ \text{is oddnumber,} \\0, & \text{if}\,\,n\ \ \text{is even number.}\end{array}\right.\) 
Since \(\alpha _{0}\ge 2,\ \ k\in \mathbb {N}\)  we obtain that \(2\alpha _{k}\ge 4,\)  for all \(k\in \mathbb {N}\)  and if we apply (REF ) we get that
\(II_{1}=0 \) 
and
\(&&II_2=\frac{1}{l_j}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\sum _{n=2^{2\alpha _k-1}}^{(j-1)/2}\frac{w_{2n+1}}{2n+1}=\frac{1}{l_j}\frac{2^{2\alpha _k(1/p-1)}r_{1}}{\Phi ^{1/2p}(2^{2\alpha _k})}\sum _{n=2^{2\alpha _k-1}}^{(j-1)/2}\frac{w_{2n}}{2n+1}.\) 
Let \(x\in I_{2}(e_{0}+e_{1}).\)  Then, by the definition of Walsh functions, we get that
\(w_{4n+2}=r_{1}w_{4n}=-w_{4n}\) 
and
\( &&\left|II_2\right|=\frac{1}{l_j}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left|\sum _{n=2^{2\alpha _k-1}}^{(j-1)/2}\frac{w_{2n}}{2n+1}\right|\) 
\(&=&\frac{1}{l_j}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left|\frac{w_{j-1}}{j}+\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4}\left(\frac{w_{4n-4}}{4n-3}+\frac{w_{4n-2}}{4n-1}\right)\right|\\ &=&\frac{1}{l_j}\frac{2^{2\alpha _{k}(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left|\frac{w_{j-1}}{j}+\sum _{n=2^{2\alpha _{k}-2}+1}^{(j-1)/4}\left( \frac{w_{4n-4}}{4n-3}-\frac{w_{4n-2}}{4n-1}\right) \right|\\ &\ge &\frac{c}{\log ({2^{2\alpha _k+1}})}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \left|\frac{w_{j-1}}{j}\right|-\sum _{n=2^{2\alpha _{k}-2}+1}^{(j-1)/4} \left|{w_{4n-4}}\right|\left( \frac{1}{4n-3}-\frac{1}{4n-1}\right) \right) \\ &&\ge \frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{1}{j}-\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4}\left( \frac{1}{4n-3}-\frac{1}{4n-1}\right)\right).\) 
By simple calculation we can conclude that
\(&&\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4}\left(\frac{1}{4n-3}- \frac{1}{4n-1}\right)=\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4}\frac{2}{(4n-3)(4n-1)} \\&\le & \sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4} \frac{2}{(4n-4)(4n-2)}=\frac{1}{2}\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4} \frac{1}{(2n-2)(2n-1)} \\&\le &\frac{1}{2}\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4} \frac{1}{(2n-2)(2n-2)}=\frac{1}{8}\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4} \frac{1}{(n-1)(n-1)} \\&\le & \frac{1}{8}\sum _{n=2^{2\alpha _k-2}+1}^{(j-1)/4} \frac{1}{(n-1)(n-2)}=\frac{1}{8}\sum _{l=2^{2\alpha _k-2}+1}^{(j-1)/4} \left(\frac{1}{n-2}-\frac{1}{n-1} \right)\\&\le &\frac{1}{8}\left(\frac{1}{2^{2\alpha _k-2}-1}-\frac{4}{j-5}\right)\le \frac{1}{8}\left(\frac{1}{2^{2\alpha _k-2}-1}-\frac{4}{j}\right).\) 
Since \(2^{2\alpha _{k}}\le j\le 2^{2\alpha _{k}+1}-1,\)  where \(\alpha _{k}\ge 2,\)  we obtain that
\(\frac{2}{2^{2\alpha _k}-4}\le \frac{2}{2^{4}-4}=\frac{1}{6}\) 
and
\( \left|II_2\right|&\ge &\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{1}{j}-\frac{1}{8}\left(\frac{1}{2^{2\alpha _k-2}-1}-\frac{4}{j}\right)\right) \\ &\ge &\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{3}{2j}-\frac{1}{2^{2\alpha _k+1}-8}\right) \\ &\ge &\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{3}{4}\frac{1}{2^{2\alpha _{k}}}-\frac{1}{2}\frac{1}{2^{2\alpha _k}-4}\right) \\ &\ge &\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{1}{4}\frac{1}{2^{2\alpha _{k}}}+\frac{1}{2}\frac{1}{2^{2\alpha _{k}}}-\frac{1}{2}\frac{1}{2^{2\alpha _k}-4}\right) \\ &=&\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{1}{4}\frac{1}{2^{2\alpha _{k}}}-\frac{2}{2^{2\alpha _k}(2^{2\alpha _k}-4)}\right)\\ &\ge &\frac{1}{4\alpha _k}\frac{2^{2\alpha _k(1/p-1)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\left( \frac{1}{4}\frac{1}{2^{2\alpha _{k}}}-\frac{1}{6}\frac{1}{2^{2\alpha _{k}}}\right)\\ &\ge &\frac{1}{48\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\ge \frac{1}{64\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}.\) 
By combining (REF ), (REF )-(REF ) for \(\in I_{2}(e_{0}+e_{1})\)  and \(0<p<1/2\)  we find that
\(&&\left|R^w_j f\left( x\right) \right|\ge \left|II_{2}\right|-\left|II_{1}\right|-\left|I \right|\\&\ge &\frac{1}{64\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}-\frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}=\frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}.\) 
Hence,
\( &&\left\Vert R^w_{j}f\right\Vert _{weak-L_p(G_2)}^p \\ &\ge &\frac{1}{128\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{1/2}(2^{2\alpha _k})}\mu \left\lbrace  x\in G_2:\left|R^w_{j}f\right|\ge \frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\right\rbrace  ^{1/p} \\ &\ge &\frac{1}{128\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{1/2}(2^{2\alpha _k})}\mu \left\lbrace  x\in I_{2}(e_{0}+e_{1}):\left|R^w_{j}f\right|\ge \frac{1}{128\alpha _k}\frac{2^{2\alpha _k(1/p-2)}}{\Phi ^{1/2p}(2^{2\alpha _k})}\right\rbrace  \\ &\ge &\frac{1}{128\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{1/2}(2^{2\alpha _k})}(\mu \left( x\in I_{2}(e_{0}+e_{1})\right))>\frac{1}{516\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{1/2}(2^{2\alpha _k})}.\) 
Moreover,
\(&&\underset{j=1}{\overset{\infty }{\sum }}\frac{\left\Vert R^w_{j}f\right\Vert _{weak-L_p(G_2)}^{p}\log ^p{(j)}\Phi (j)}{j^{2-2p} }\) 
\(&\ge & \underset{\left\lbrace  j\in \mathbb {A}_{0,2}:\text{ }2^{2\alpha _k}< j\le 2^{2\alpha _{k}+1}-1\right\rbrace  }{\sum }\frac{\left\Vert R^w_{j}f\right\Vert _{weak-L_p}^{p}\log ^p{(j)}\Phi (j)}{j^{2-2p} } \\&\ge &\frac{c}{\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{p/2}(2^{2\alpha _k)}}\underset{\left\lbrace  j\in \mathbb {A}_{0,2}:\text{ }2^{2\alpha _k}< j\le 2^{2\alpha _{k}+1}-1\right\rbrace  }{\sum }\frac{\log ^p{(j)}\Phi (j)}{j^{2-2p}}\\&\ge &\frac{c\Phi (2^{2\alpha _{k}})\log ^p{(2^{2\alpha _k})}}{\alpha _k^p}\frac{2^{2\alpha _k(1-2p)}}{\Phi ^{1/2}(2^{2\alpha _k})}\underset{\left\lbrace  j\in \mathbb {A}_{0,2}:\text{ }2^{2\alpha _k}< j\le 2^{2\alpha _{k}+1}-1\right\rbrace  }{\sum }\frac{1}{j^{2-2p}}\\&\ge & \Phi ^{1/2}(2^{2\alpha _{k}})\rightarrow \infty ,\text{ \ as \ \ }k\rightarrow \infty .\) 
The proof is complete.
